Chapter 9, Problem 12DQ

(a)

Interpretation Introduction

Interpretation: The value of change in enthalpy, internal energy, heat and work done needs to be compared for two paths mentioned in the question.

Concept Introduction:

For a process at constant pressure, the work dome can be calculated as follows:

  $w=-P\Delta V$

Also, value of heat, q can be calculated as follows:

  $q_P=n{C}_P\Delta T$

The change in enthalpy is equal to $q_P$ for constant pressure condition. The change in internal energy can be calculated s follows:

  $\Delta E=q+w$

Here, q is heat and w is work done.

Explanation of Solution

The initial and final values of pressure and volume are as follows:

  $\begin{array}{l}{P}_1=1\text{ atm}\\ P_2=2\text{ atm}\\ V_1=1.0\text{ L}\\ V_2=2.0\text{ L}\end{array}$

Since, the changes occur in path then there are two paths possible for the changes.

In path 1: Initially, 1 L volume changes to 2 L and pressure remains the same that is 1 atm after that pressure changes from 1 atm to 2 atm and volume is 2 L.

In path 2: Initially, 1 atm pressure changes to 2 atm and volume remains the same that is 1 L. After that volume changes from 1 L to 2 L and pressure is 2 atm.

The enthalpy change and internal energy change does not depend on paths as they are state functions. Thus, for path 1 and path 2, the change in enthalpy and internal energy remains the same.

Heat and work are path functions because the values depend on path. Thus, values of q and w will not be same for path 1 and path 2.

In path 1, gas expands and change in volume takes place from 1 L to 2 L against a constant pressure of 1 atm but in second path, the expansion takes place against pressure 2 atm.

Due to the above reason, the work done in first path will be less than the second path. According to first law of thermodynamics, the change in internal energy, heat and work done are related to each other as follows:

  $\Delta E=q+w$

Here, the value of change in internal energy is equal in both the paths and the value of w is higher in path 2 thus, the value of q should be lower in path 2.

(b)

Interpretation Introduction

Interpretation: The change in internal energy, change in enthalpy, heat and work done needs to be calculated for each case and if the calculation makes sense or not needs to be explained.

Concept Introduction:

For a process at constant pressure, the work dome can be calculated as follows:

  $w=-P\Delta V$

Also, value of heat, q can be calculated as follows:

  $q_P=n{C}_P\Delta T$

The change in enthalpy is equal to $q_P$ for constant pressure condition. The change in internal energy can be calculated s follows:

  $\Delta E=q+w$

Here, q is heat and w is work done.

Explanation of Solution

If path 1 is considered, there are mainly 2 steps. In step 1, pressure is constant and change in volume from 1 L to 2 L takes place.

The work done for this step can be calculated as follows:

  $w_1=-P\Delta V$

Putting the values,

  $\begin{array}{l}{w}_1=-\left(1.0\text{ atm}\right)\left(2-1\right)\text{ L}\\ \text{=1}\text{.0 L atm}\left(\frac{101.325\text{ J}}{\text{1 L atm}}\right)\\ =-101\text{ J}\end{array}$

Using the ideal gas equation,

  $\begin{array}{l}P\Delta V=nR\Delta T\\ \Delta T=\frac{P\Delta V}{nR}\\ =\frac{1.01\text{ J}}{nR}\end{array}$

At constant pressure,

  $q_1=\Delta H=n{C}_P\Delta T$

Putting the values,

  $q_1=\Delta H=n{C}_P\Delta T$

Putting the values,

  $q_1=\Delta H=n\left(\frac{5}{2}R\right)\left(\frac{101\text{ J}}{nR}\right)=253\text{ J}$

The value of change in internal energy can be calculated as follows:

  $\begin{array}{l}\Delta E_1=n{C}_V\Delta T\\ =n\left(\frac{3}{2}R\right)\left(\frac{101\text{ J}}{nR}\right)\\ =152\text{ J}\end{array}$

Now, for step 2 in path 1, volume is constant and change in pressure takes place from 1 atm to 2 atm.

Since, the change in volume does not take place; the work done will be zero.

From the ideal gas law, the change in T can be calculated as follows:

  $\Delta T=\frac{\Delta PV}{nR}$

Putting the values,

  $\begin{array}{l}\Delta T=\frac{\left(2-1\right)\text{ atm}\left(2\text{ L}\right)}{nR}\\ =\frac{\left(2\text{ L atm}\right)}{nR}\left(\frac{101.325\text{ J}}{1\text{ L atm}}\right)\\ =\frac{203\text{ J}}{nR}\end{array}$

The value of change in internal energy will be equal to $q_2$

Thus,

  $\begin{array}{l}{q}_2=\Delta E_2=n{C}_V\Delta T\\ =n\left(\frac{3}{2}R\right)\left(\frac{203\text{ J}}{nR}\right)\\ =304\text{ J}\end{array}$

Similarly,

  $\begin{array}{l}\Delta H_2=n{C}_P\Delta T\\ =n\left(\frac{5}{2}R\right)\left(\frac{203\text{ J}}{nR}\right)\\ =508\text{ J}\end{array}$

The total value of work done, heat, change in internal energy and change in enthalpy for first path will be:

  $\begin{array}{l}q=q_1+q_2\\ =\left(253+304\right)\text{ J}\\ =557\text{ J}\end{array}$

Similarly,

  $\begin{array}{l}w=w_1+w_2\\ =\left(-101+0\right)\text{ J}\\ =-101\text{ J}\end{array}$

Also,

  $\begin{array}{l}\Delta E=\Delta E_1+\Delta E_2\\ =\left(152+304\right)\text{ J}\\ \text{=456 J}\end{array}$

And,

  $\begin{array}{l}\Delta H=\Delta H_1+\Delta H_2\\ =\left(253+508\right)\text{ J}\\ \text{=761 J}\end{array}$

Now, considering path 2 there are also two parts.

In step 1 of the path 2:

Volume is constant and change in pressure takes place from 1 to 2 atm.

Thus, work done will be 0.

The change in temperature can be calculated as follows:

  $\Delta T=\frac{\Delta PV}{nR}$

Putting the values,

  $\begin{array}{l}\Delta T=\frac{\left(2-1\right)\text{ atm}\left(\text{1 L}\right)}{nR}\\ =\frac{\left(1\text{ L atm}\right)}{nR}\left(\frac{101.325\text{ J}}{1\text{ L atm}}\right)\\ =\frac{101.325\text{ J}}{nR}\end{array}$

The value of change in internal energy will be equal to $q_3$

Thus,

  $\begin{array}{l}{q}_3=\Delta E_3=n{C}_V\Delta T\\ =n\left(\frac{3}{2}R\right)\left(\frac{\text{101}\text{.325 J}}{nR}\right)\\ =152\text{ J}\end{array}$

Similarly,

  $\begin{array}{l}\Delta H_3=n{C}_P\Delta T\\ =n\left(\frac{5}{2}R\right)\left(\frac{101.325\text{ J}}{nR}\right)\\ =253\text{ J}\end{array}$

Now, for step 2 of path 2:

The change in volume takes place from 1 to 2 L at constant pressure thus, work done will be:

  $\begin{array}{l}{w}_4=-P\Delta V\\ =-\left(2\text{ L atm}\right)\left(\frac{101.325\text{ J}}{1\text{ L atm}}\right)\\ =-203\text{ J}\end{array}$

The change in temperature can be calculated as follows:

  $\begin{array}{l}\Delta T=\frac{\Delta PV}{nR}\\ =\frac{203\text{ J}}{nR}\end{array}$

The value of change in internal energy will be equal to q₄:

  $\begin{array}{l}{q}_4=\Delta H_4=n{C}_P\Delta T\\ =n\left(\frac{5}{2}R\right)\left(\frac{203\text{ J}}{nR}\right)\\ =508\text{ J}\end{array}$

Similarly,

  $\begin{array}{l}\Delta E_4=n{C}_V\Delta T\\ =n\left(\frac{3}{2}R\right)\left(\frac{202.65\text{ J}}{nR}\right)\\ =304\text{ J}\end{array}$

The total value of work done, heat, change in internal energy and change in enthalpy for second path will be:

  $\begin{array}{l}q=q_3+q_4\\ =\left(152+508\right)\text{ J}\\ =660\text{ J}\end{array}$

Similarly,

  $\begin{array}{l}w=w_3+w_4\\ =\left(0+\left(-204\right)\right)\text{ J}\\ =-204\text{ J}\end{array}$

Also,

  $\begin{array}{l}\Delta E=\Delta E_3+\Delta E_4\\ =\left(152+304\right)\text{ J}\\ \text{=456 J}\end{array}$

And,

  $\begin{array}{l}\Delta H=\Delta H_3+\Delta H_4\\ =\left(253+508\right)\text{ J}\\ \text{=761 J}\end{array}$

Since, w and q are state functions thus, values for path 1 and path 2 are not same but sum of heat and work done is equal to change in internal energy in path 1 as well as path 2 this is in consistent with 1^st law of thermodynamics.

The value of change in enthalpy and change in internal energy are state functions. The values are same for both paths.